Lemma 69.9.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. If $X$ is separated, locally Noetherian, and the dimension of the local ring of $X$ at $x$ is $\leq 1$ (Properties of Spaces, Definition 63.10.2), then there exists an open subspace of $X$ containing $x$ which is a scheme.

** Separated algebraic spaces are schemes in codimension 1. **

**Proof.**
(Please see the remark below for a different approach avoiding the material on finite groupoids.) We can replace $X$ by an quasi-compact neighbourhood of $x$, hence we may assume $X$ is quasi-compact, separated, and Noetherian. There exists a scheme $U$ and a finite surjective morphism $U \to X$, see Limits of Spaces, Proposition 67.16.1. Let $R = U \times _ X U$. Then $j : R \to U \times _ S U$ is an equivalence relation and we obtain a groupoid scheme $(U, R, s, t, c)$ over $S$ with $s, t$ finite and $U$ Noetherian and separated. Let $\{ u_1, \ldots , u_ n\} \subset U$ be the set of points mapping to $x$. Then $\dim (\mathcal{O}_{U, u_ i}) \leq 1$ by Decent Spaces, Lemma 65.12.6.

By More on Groupoids, Lemma 40.14.10 there exists an $R$-invariant affine open $W \subset U$ containing the orbit $\{ u_1, \ldots , u_ n\} $. Since $U \to X$ is finite surjective the continuous map $|U| \to |X|$ is closed surjective, hence submersive by Topology, Lemma 5.6.5. Thus $f(W)$ is open and there is an open subspace $X' \subset X$ with $f : W \to X'$ a surjective finite morphism. Then $X'$ is an affine scheme by Cohomology of Spaces, Lemma 66.17.1 and the proof is finished. $\square$

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